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Here is a proof that $a_1(n, p, q) = a_2(n,p,q)$. The generating function for the Stirling numbers of the second kind is $$\sum_{n=i}^\infty {n\brace i}\frac{x^n}{n!}= \frac{(e^x-1)^i}{i!}.$$ Also $$\ …

Something much more general is true. Let $p$ be a prime. All congruences in what follows are modulo $p$. A Hurwitz series is a power series of the form $\sum_{n=0}^\infty a_n z^n/n!$ where the $a_n$ …

This is a known result. To quote from Richard Stanley's Enumerative Combinatorics, Volume 2, second edition, solution to problem 8(e) of Chapter 5, page 115: This is equivalent to a conjecture of J. M …

We proceed by "guessing" a generating function for $R(n,q)$ and verifying that it has the right properties. According to https://oeis.org/A143017, the generating function $G= \sum_{n=1}^\infty a(n) x^ …

Let \begin{equation*} A(x,q) = e^{qx}A(x)=e^{(q+1)x+(e^x-1)^2/2} \end{equation*} and define $a(n,q)$ by \begin{equation*} A(x,q) = \sum_{n=0}^\infty a(n,q) \frac{x^n}{n!}. \end{equation*} Then $a(n,0) …

Let $$ A(x,q)=\sum_{i=0}^{\infty}\frac{x^i}{\prod\limits_{j=0}^{i}(1-f(q+j)x)}, $$ so that $A(x) = A(x,0)$. Define $a(n,q)$ by $$A(x,q) = \sum_{n=0}^\infty a(n,q) x^n,$$ so that $a(n) = a(n,0)$. I wil …

This product appears in permutation enumeration. See, for example, D. P. Roselle, Coefficients associated with the expansion of certain products (DOI), Proc. Amer. Math. Soc. 45 (1974), 144-150.

No. It's easy to see that $B_n=4$ is impossible, and $B_n$ is never divisible by 8. This follows from the fact that $B_n$ is periodic modulo 8 with period 24. See, e.g., W. F. Lunnon, P. A. B. Pleasan …

More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity. Similarly, $$e^x\sum_{k=0}^\infty F_{n+k …

A comprehensive account of these congruences can be found in Romeo Meštrović,Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862–2012), https://doi.org …

Glaisher's I-numbers are described in J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., 31 (1899), 216-235.

Here's a straightforward proof, using generating functions, though it's not as elegant as Ofir's. We have $$ \sum_{a=0}^\infty\binom{a}{j}x^a =\frac{x^j}{(1-x)^{j+1}}. $$ Setting $j=(p-1)k$ and summin …

The sum can be expressed in terms of hypergeometric series as $$(n+1)\binom{2n}{n}\binom{2n+1}{n}\,{}_3F_2\left({-n,\,\tfrac12,\,\tfrac12\atop -n+\tfrac12,\tfrac32}\biggm| 1\right).$$ This means that …

Is the following true? Let $p$ be a prime and let $w$ be a $(p-1)$st root of unity (not necessarily primitive). Then $$\binom{w}{n}=\frac{w(w-1)\cdots(w-n+1)}{n!}$$ is $p$-integral; i.e., it can be ex …

Although this isn't an answer to the question, it's worth pointing out that the second and third families are essentially binomial coefficients. We have $$U_2(m,n):=\frac{(2m)!\,(2n)!}{m!\, n!\, (m+n) …